CBSE Class 9 Answered

the satellite of mass 1000 kg is supposed to orbit the earth at height of 2000 km about the Earth surface. the kinetic energy and potential energy respectively are

Asked by ambica.mallarapu | 30 Oct, 2023, 09:58: PM

Expert Answer

Radius R of earth = 6.371 × 10⁶ m

If the satellite is at a distance h = 2000 km above earth surface , then centripetal force is calculated as

begin mathsize 14px style m space fraction numerator v squared over denominator R plus h end fraction space equals space G space fraction numerator M space space m over denominator left parenthesis R plus h right parenthesis squared end fraction end style

where m is mass of satellite , v is speed of satellite in its orbit , G is universal gravitational constant and

M is mass of earth

from above expression , we get

begin mathsize 14px style v space equals space square root of fraction numerator G space M over denominator left parenthesis R plus h right parenthesis end fraction end root space equals space square root of fraction numerator 6.674 space cross times 10 to the power of negative 11 end exponent cross times 5.972 cross times 10 to the power of 24 over denominator 8.371 cross times 10 to the power of 6 end fraction end root space equals space 6900 space m divided by s end style

Kinetic energy K of satellite,

begin mathsize 14px style K space equals space left parenthesis 1 divided by 2 right parenthesis space m space v squared space equals space 0.5 space cross times space 1000 space cross times space left parenthesis 6900 right parenthesis squared space equals space 2.381 space cross times space 10 to the power of 10 space J end style

Potential energy U of satellite ,

begin mathsize 14px style U space equals space minus space G space fraction numerator M space m over denominator left parenthesis R plus h right parenthesis end fraction space equals space fraction numerator negative space 6.674 space cross times space 10 to the power of negative 11 end exponent space cross times 5.972 space cross times space 10 to the power of 24 over denominator 8.371 space cross times space 10 to the power of 6 end fraction space equals space minus 4.761 space cross times space 10 to the power of 7 space J end style

Answered by Thiyagarajan K | 31 Oct, 2023, 11:01: AM

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